What is the difference between the non-minimal coupling of GR to a scalar field and the coupling of the metric to the stress tensor?

Question

In General Relativity we can non-minimally couple the action of the gravitational field to a scalar field, as seen here, where the lowest order gravitational action is coupled to the scalar fields with the form: $\kappa R\phi^2$.

In linearized gravity it is assumed that the gravitational field couples to the stress tensor and not directly to the field $\phi$, with the form $\kappa h_{\mu\nu}T^{\mu\nu}$, as seen here.

What is the difference between coupling directly to the field $\phi$ vs the stress tensor $T^{\mu\nu}$?

In the linearized case, why do we not take the coupling directly with $\phi$ as $\kappa h\phi ^2$, where $h=h^{\mu}_{\mu}=\eta^{\mu\nu}h_{\mu\nu}$ instead?

Answer 1

Since $\frac{\partial R}{\partial g_{\mu\nu}} \neq 0$, the Hilbert procedure yields a stress tensor for a scalar field which depends on the coefficient of $R\phi^2$. This is the only thing that makes the two cases differ.

The minimally coupled system linearizes to an action where $h_{\mu\nu}$ multiplies the stress tensor for a scalar without $R\phi^2$. The non-minimally coupled system linearizes to an action where $h_{\mu\nu}$ multiplies the stress tensor for a scalar with $R\phi^2$.

It's true that $R\phi^2$ is not the only way of making the coupling between a scalar and gravity non-minimal. But every time I see one of these being used, the combination of $\phi$ and curvature invariants is written explicitly. If an author actually wrote \begin{equation} \int \left ( \frac{1}{2\kappa} R + \mathcal{L}_M + g_{\mu\nu} T^{\mu\nu} \right ) \sqrt{-g} \, dx \end{equation} where $T^{\mu\nu}$ is the thing that would've been the matter stress tensor before $g_{\mu\nu} T^{\mu\nu}$ was added, this would be very confusing. Because this would be equivalent to just rescaling the potential in $\mathcal{L}_M$.